What is the solution for 2x^3y^'=y(y^2+3x^2) ?

The solution for 2x^3y^'=y(y^2+3x^2) is y=(xsqrt(x))/(sqrt(-x+c_{1))},y=-(xsqrt(x))/(sqrt(-x+c_{1))}

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2x^3y^'=y(y^2+3x^2)

2 x^{3} y^{^{'}} = y (y^{2} + 3 x^{2})

y = \frac{x x}{- x + c _{1}}, y = - \frac{x x}{- x + c _{1}}

Solution steps

2 x^{3} y^{'} (x) = y (y^{2} + 3 x^{2})

Solve Bernoulli ODE:

y = \frac{x x}{- x + c _{1}}, y = - \frac{x x}{- x + c _{1}}

y = \frac{x x}{- x + c _{1}}, y = - \frac{x x}{- x + c _{1}}

\int 6 x + 1 d x

y = - \frac{x ^{2}}{40}, (- 45, - 2)

\frac{d}{d x} (f (x) (2 x^{2} - 4))

4 x, x^{3}

\frac{\partial}{\partial x} ((x + 1)^{2})

What is the solution for 2x^3y^'=y(y^2+3x^2) ?
The solution for 2x^3y^'=y(y^2+3x^2) is y=(xsqrt(x))/(sqrt(-x+c_{1))},y=-(xsqrt(x))/(sqrt(-x+c_{1))}